Evaluate the limit $\lim_{x \to 0+}\int_{x}^{2x}\frac{\sin^mt}{t^n}dt$

Question

Evaluate the limit $$\lim_{x \to 0+}\int_{x}^{2x}\frac{\sin^mt}{t^n}dt$$ where m ,n are natural numbers. Take cases when $m\ge n$,,when $n-m=1$,when $n-m>1$

I started out by multipying numerator and denominator by x and then used L-hospitals rule, but it became very ugly. I was left without options as problems of this type usually could be solved by this method.So here are somethings i tried to use :

when $m=n$

$$\lim_{t \to 0}\frac{\sin^mt}{t^n}=1$$

when $m>n$
$$\lim_{t \to 0}\frac{\sin^mt}{t^n}=0$$
when $m<n$:

$\lim_{t \to 0}\frac{\sin^mt}{t^n}$ is not defined

I am looking for a solution that uses elementary methods. Somethings i already know are leibnitz rule,squeeze play theorem

Answer 1

Just an idea, that I suppose can be fruitful.
By the mean value theorem, $\exists \xi\in(x,2x)$ such that $$ \int_x^{2x}\frac{(\sin t)^m}{t^n}dt=x \frac{(\sin \xi)^m}{\xi^n} $$ moreover $$ \frac{1}{2^nx^n}\leq\frac{1}{\xi^n}\leq\frac{1}{x^n} $$ and for small $x,$ such that $0<x<\pi/2$ $$ (\sin x)^m\leq(\sin\xi)^m\leq[\sin(2x)]^m $$ It suggests, by the squeeze theorem, that the result is $0$ for $m\geq n$ and $+\infty$ for $n-m>1$, but fails to give information for the most interesting case $n-m=1.$

Edit

Following the suggestion of @J.G. (see comment to the question) $$ ct\leq \sin t\leq t, \qquad \forall t\in[x,2x] $$ so that $$ c^mt^{m-n}=\frac{(ct)^m}{t^n}\leq\frac{(\sin t)^m}{t^n}\leq\frac{t^m}{t^n}=t^{m-n} $$ The integral becomes immediate: if $m-n\neq-1$ $$ c^m\left.\frac{t^{m-n+1}}{m-n+1}\right|_x^{2x}=\int_x^{2x} c^mt^{m-n}dt\leq\int_x^{2x}\frac{(\sin t)^m}{t^n}dt\leq\int_x^{2x} t^{m-n}dt=\left.\frac{t^{m-n+1}}{m-n+1}\right|_x^{2x} $$ and finally $$ c^m\frac{2^{m-n+1}-1}{m-n+1}x^{m-n+1}\leq\int_x^{2x}\frac{(\sin t)^m}{t^n}dt\leq\frac{2^{m-n+1}-1}{m-n+1}x^{m-n+1} $$ If $m-n>-1$ them $\lim_{x\to0} x^{m-n+1}=0$ and the integral is $0$.
If $m-n<-1$ them $\lim_{x\to0^+} x^{m-n+1}=+\infty$ and the coefficient $$ \frac{2^{m-n+1}-1}{m-n+1} $$ is positive, and so the integrale is $+\infty.$
Finally, if $m-n=-1,$ then $$ c^m\frac{1}{t}\leq\frac{(\sin t)^m}{t^n}\leq\frac{1}{t} $$ and $$ c^m\ln 2=c^m\left.\ln t\right|_x^{2x}=\int_x^{2x}c^m\frac{1}{t}dt\leq\int_x^{2x}\frac{(\sin t)^m}{t^n}dt\leq\int_x^{2x}\frac{1}{t}dt=\left.\ln t\right|_x^{2x}=\ln 2 $$ and given that $c\to1$ when $x\to0^+$, the integral evaluate to $\ln 2$.